Let $G$ be a finite abelian group. We denote by $Z^3(G,\mathbb C^{\ast})$ the group of normalized $3$-cocycles. For any $\omega\in Z^3(G,\mathbb C^{\ast})$ and $g\in G$ we have the map $$\omega_g (x,y)=\frac{\omega(g,x,y)\omega(x,y,g)}{\omega(x,g,y)}$$ Now let $Z^3(G,\mathbb C^{\ast})_{ab}$ denote the set of all normalized $3$-cocycles $\omega$ for which $\omega_g$ is a $2$-coboundary for all $g\in G$, and let $H^3(G,\mathbb C^{\ast}) _{ab}$ be the corresponding set of cohomology classes. It is not hard to check that $H^3(G,\mathbb C^{\ast})_{ab}$ is a subgroup of $H^3(G,\mathbb C^{\ast})$, and your question is asking for an example when it is a proper subgroup.
An easier way to do this is to look for a different description of $H^3(G,\mathbb C^{\ast})_{ab}$. It is not hard to check that $\omega_{g}(x,y)=\omega_g(y,x)$ for all $(x,y)\in G\times G$ iff $\omega_g$ is a $2$-coboundary. So if you define the map $\psi^{\ast}: H^3(G,\mathbb C^{\ast})\to Hom(\bigwedge^3 G,\mathbb C^{\ast})$ $$\psi^{\ast}([\omega])(x,y,z)=\frac{\omega(x,y,z)\omega(y,z,x)\omega(z,x,y)}{\omega(y,x,z)\omega(z,y,x)\omega(x,z,y)}$$ $\psi^{\ast}([\omega])(x,y,z)=\frac{\omega(x,y,z)\omega(y,z,x)\omega(z,x,y)}{\omega(y,x,z)\omega(z,y,x)\omega(x,z,y)}=\frac{\omega_z(x,y)}{\omega_z(y,x)}$$then H^3(G,\mathbb C^{\ast})_{ab} is precisely the kernel of \psi^{\ast} (Lemma 7.4 in the paper above). Now \psi^{\ast} is surjective so the question becomes: when is Hom(\bigwedge^3 G,\mathbb C^{\ast}) non-trivial? This is the case whenever G is the direct sum of at least 3 cyclic factors, in particular any (\mathbb Z/n\mathbb Z)^3 works. An explicit example in this case is \omega sending (x,y,z) \omega(x,y,z)=\mu^{x_1y_2z_3} where x=(x_1,x_2,x_3) etc. to \mu^{x_1y_2z_3}, where and \mu is a primitive nth root of unity. Then we have \psi^{\ast}([\omega])(x,y,z)=\mu^{\det(x,y,z)} which is non-trivial. In particular \omega_x is non-trivial in H^2. 1 Following up on what has already been said, this is an explanation taken from "Group cohomology and gauge equivalence of some twisted quantum doubles", by Geoffrey Mason and Siu-Hung Ng. Let G be a finite abelian group. We denote by Z^3(G,\mathbb C^{\ast}) the group of normalized 3-cocycles. For any \omega\in Z^3(G,\mathbb C^{\ast}) and g\in G we have the map$$\omega_g (x,y)=\frac{\omega(g,x,y)\omega(x,y,g)}{\omega(x,g,y)}$$Now let Z^3(G,\mathbb C^{\ast})_{ab} denote the set of all normalized 3-cocycles \omega for which \omega_g is a 2-coboundary for all g\in G, and let H^3(G,\mathbb C^{\ast}) _{ab} be the corresponding set of cohomology classes. It is not hard to check that H^3(G,\mathbb C^{\ast})_{ab} is a subgroup of H^3(G,\mathbb C^{\ast}), and your question is asking for an example when it is a proper subgroup. An easier way to do this is to look for a different description of H^3(G,\mathbb C^{\ast})_{ab}. It is not hard to check that \omega_{g}(x,y)=\omega_g(y,x) for all (x,y)\in G\times G iff \omega_g is a 2-coboundary. So if you define the map \psi^{\ast}: H^3(G,\mathbb C^{\ast})\to Hom(\bigwedge^3 G,\mathbb C^{\ast})$$\psi^{\ast}([\omega])(x,y,z)=\frac{\omega(x,y,z)\omega(y,z,x)\omega(z,x,y)}{\omega(y,x,z)\omega(z,y,x)\omega(x,z,y)}$$then $H^3(G,\mathbb C^{\ast})_{ab}$ is precisely the kernel of$\psi^{\ast}$(Lemma 7.4 in the paper above). Now$\psi^{\ast}$is surjective so the question becomes: when is$Hom(\bigwedge^3 G,\mathbb C^{\ast})$non-trivial? This is the case whenever$G$is the direct sum of at least$3$cyclic factors, in particular any$(\mathbb Z/n\mathbb Z)^3$works. An explicit example in this case is$\omega$sending$(x,y,z)$where$x=(x_1,x_2,x_3)$etc. to$\mu^{x_1y_2z_3}$, where$\mu$is a primitive$n$th root of unity. Then we have$\psi^{\ast}([\omega])(x,y,z)=\mu^{\det(x,y,z)}\$ which is non-trivial.